Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 (Causal Inference)

Under complete randomization, every assignment vector with exactly $n_z$ patients receiving treatment $z$ is equally likely. Thus, for $z=(z_1,\dots ,z_n)$ and $N_r(z)={\sum }_i{1}_{\{z_i=r\}}$,The reciprocal probability $n!/(n_0!n_1!n_2!)$ is the multinomial coefficient counting assignments with those arm sizes.

Under independent assignment,The arm-count vector has a multinomial distribution. Conditional on $N_r=n_r$ for $r=0,1,2$, every compatible vector has the same factor ${\prod }_r{\pi }_r^{n_r}$, sofor compatible $z$, and zero otherwise. Conditioning independent assignment on its arm sizes therefore recovers complete randomization.

For a two-by-two contingency table, Fisher's exact test conditions on both row totals and both column totals. Under the null of no association, the upper-left count $K$ then has the hypergeometric distributionwhere $r_1$ and $c_1$ are the first row and column totals. A one-sided p-value sums the appropriate hypergeometric tail; a common two-sided p-value sums the probabilities of all feasible tables no more likely under the null than the observed table. Because the conditional distribution is discrete, the p-value is super-uniform random variable rather than generally exactly uniform.

Cross-classify the binary exposure $A_i=1_{\{Z_i\ge 1\}}$ against the binary outcome $Y_i$. The four cells count placebo survivors or deaths and active-treatment survivors or deaths.

Under the sharp causal null hypothesis, every observed outcome equals the fixed value $Y_i(0)=Y_i(1)=Y_i(2)$, regardless of assignment. With fixed arm sizes, the active set is a uniformly chosen subset of size $n_1+n_2$. Under independent assignment, $A_i$ are independent Bernoulli variables with probability ${\pi }_1+{\pi }_2$, and conditioning on their total again makes the active set uniform. Conditional on the table margins, the active-group outcome count therefore has exactly the hypergeometric distribution used by Fisher's exact test.

The conditional p-value is super-uniform for every set of margins. The law of total probability then givesHence the test is valid under either assignment mechanism.

Within the active set, cross-classify dosage $Z_i\in \{1,2\}$ against $Y_i$. Conditional on the vector $A=(A_1,\dots ,A_n)$, complete randomization assigns $n_1$ of the active patients to low dosage and $n_2$ to high dosage uniformly. Under independent assignment, active patients receive low and high dosage with conditional probabilitiesconditioning further on their low- and high-dose totals again gives the same uniform allocation. Under $H_B$, active patients' outcomes are fixed as their common $Y_i(1)=Y_i(2)$, so the conditional Fisher p-value obeys

Under $H_A$, $P_A$ is a function only of $A$ and the fixed outcomes. Applying the law of iterated expectation under the joint null givesThis conditional argument explains the stated near-independence even though the two tests reuse outcomes.

A positive density factorizes according to an undirected graphical model $G^u=(\mathcal{V},\mathcal{E})$ when there are nonnegative clique potentials ${\psi }_C$ such thatwhere $\mathcal{C}$ may be taken as the maximal cliques and $Z$ normalizes the density. Equivalently, each potential involves only variables in one complete subgraph.

The global Markov property for an undirected graph says that whenever a vertex set $S$ separates disjoint sets $A$ and $B$ in the graph,The Hammersley-Clifford theorem states that a strictly positive density factorizes over the cliques of an undirected graph if and only if it satisfies this global Markov property. Strict positivity is essential for the converse from conditional independences to factorization.

Let $K={\Sigma }^{-1}$ be the precision matrix, let $R=\{1,\dots ,p\}\setminus \{j,k\}$, and condition on $V_R=v_R$. In the Gaussian exponent, all terms depending jointly on $v_j$ and $v_k$ are contained inIf $K_{jk}=0$, this conditional density is a product of one function of $v_j$ and one function of $v_k$, so the conditional independence of $V_j$ and $V_k$ given $V_R$ holds.

Conversely, conditional independence makes this everywhere-positive conditional density factorize. Its mixed second derivative must therefore vanish:Hence the conditional independence holds exactly when $({\Sigma }^{-1}{)}_{jk}=0$.

A density factorizes according to a Directed acyclic graph whenwhere $\mathrm{pa}(j)$ is the set of parents of vertex $j$.

The required undirected graph is the moral graph: join every pair of parents having a common child, retain the parent-child adjacencies, and remove all arrowheads. Each DAG family $\{j\}\cup \mathrm{pa}(j)$ is then a clique, so each conditional factor is a clique potential and the DAG factorization is also an undirected factorization. These edges are minimal for a guarantee covering every DAG-factorizing density, because an arbitrary conditional factor can couple every pair of variables in its family.

A nonparametric structural equation model for the graph can be writtenThe bidirected edge $A\leftrightarrow Y$ permits dependence between $U_A$ and $U_Y$; apart from this pair the exogenous variables are mutually independent. The basic potential outcomes areand the natural nested outcome is $Y(a)=Y(M(a,X),X)$.

The exogenous independences imply that the whole family $\{M(a,x):a,x\}$ is independent of $\{X,A,Y(m,x):m,x\}$, while $X$ is independent of $A$ and the basic response potentials. The bidirected edge means that $A$ and $Y(m,x)$ need not be independent. Useful observed-counterfactual consequences include

In an acyclic directed mixed graph, a fixable vertex $v$ is one whose bidirected district contains no proper directed descendant of $v$:Here $A$ and $Y$ lie in the same district because $A\leftrightarrow Y$, and $Y$ is a directed descendant of $A$ along $A\to M\to Y$. Thus $A$ is not fixable.

The claim is not implied. On the path $A\leftrightarrow Y\leftarrow X$, $Y$ is a collider, and conditioning on $Y$ opens that path. Equivalently, conditioning on the common effect $Y$ can induce collider bias between $A$ and $X$.

The claim is not implied. In the graph for $Y(m)$ the pathis open after conditioning only on $X$. The latent common cause represented by the bidirected edge associates $A$ with $Y(m)$, and $A$ also causes $M$.

The claim is not implied. The intervention setting $A=a$ removes the outgoing dependence of descendants on the observed value of $A$, but the bidirected path $A\leftrightarrow Y(a)$ remains open. Conditioning on $X$ does not block latent confounding between $A$ and $Y$.

This conditional independence is implied. Fixing the mediator at $m$ removes the directed edge $M\to Y$ in the relevant counterfactual graph. Conditioning on $A$ blocks $M\leftarrow A\leftrightarrow Y(m)$, while conditioning on $X$ blocks $M\leftarrow X\to Y(m)$. Every remaining path is blocked by the m-separation criterion, so

This conditional independence is implied. In the graph for $M(a)$, the incoming causal value of $A$ has been fixed, so the path $A\to M$ no longer transmits association to $M(a)$. The only possible route through $Y$ contains a collider, and conditioning on $X$ blocks the common-cause path through $X$. Hence m-separation gives

This is the covariate-conditional front-door adjustment. The paths from $A$ to $M$ have no unblocked backdoor path after conditioning on $X$, and all backdoor paths from $M$ to $Y$ are blocked by $(A,X)$. ThereforeThe inner sum identifies the effect of setting $M=m$ at covariate value $x$ by adjusting for $A$; the outer sum transports this through the mediator distribution generated by setting $A=a$ and then averages over $X$.

For binary $A$, conditional covariance satisfiesBy conditional exchangeability and consistency of potential outcomes, the difference in braces is $\mathbb{E}[Y(1)-Y(0)\mid X]$. Consequentlyso the required weight isThis overlap weight emphasizes covariate strata with treatment probabilities near one half and downweights strata near a violation of positivity in causal inference.

Write $p_x=\mathbb{P}(X=x)$, ${\pi }_x=\mathbb{E}(A\mid X=x)$, and ${\mu }_x=\mathbb{E}(Y\mid X=x)$. The functional isThe first term has influence function $AY-\mathbb{E}(AY)$. Applying the product rule to the second term, including perturbations of $p_x$, ${\pi }_x$, and ${\mu }_x$, givesSubtracting and using $\beta =\mathbb{E}(AY)-\mathbb{E}[\pi \mu ]$ yieldsIts expectation is zero because the first product has expectation $\mathbb{E}\{\mathrm{Cov}(A,Y\mid X)\}=\beta$.

Condition on the independent external sample, and abbreviate ${\delta }_{\pi }={\hat{\pi }}_m-\pi$ and ${\delta }_{\mu }={\hat{\mu }}_m-\mu$. Expanding the plug-in estimator around the true nuisance functions giveswhere $\sqrt{n}{R}_{n,m}=o_p(1)$ because the two nuisance mean squared errors vanish, the evaluation sample is independent of the nuisance fits, $A$ is bounded, and $Y$ has bounded support. The linear term is the empirical average of the influence function, so the central limit theorem givesThe remaining bias obeys the Cauchy-Schwarz inequalityThus the claimed conclusion follows under the standard product-rate conditionequivalently $\sqrt{n}\sqrt{\mathrm{MSE}({\hat{\pi }}_m)operatornameMSE({\hat{\mu }}_m)}\to 0$. Slutsky theorem then yields

As printed, the paper instead assumes only $\sqrt{n}\mathrm{MSE}({\hat{\pi }}_m)\mathrm{MSE}({\hat{\mu }}_m)\to 0$, which is insufficient with its stated definition of MSE. For example, deterministic nuisance errors of size $n^{-3/16}$ have both MSEs equal to $n^{-3/8}$ and satisfy the printed condition, while the scaled product bias is $\sqrt{n}{n}^{-3/8}=n^{1/8}$. The result therefore requires the stronger condition above, or “MSE” in the printed rate must be read as root mean squared error.